Let $S$ be some (fixed) subset of $\mathbb{Z} [X_1, \dots , X_n]$ which contains only homogeneous polynomials, and if $F$ is a field, let $X(F)$ be the set of $ x \in P^{n-1}(F)$ such that $f (x) = 0$ for any $f \in S$.

Now consider the assertion $(E_p) : X(F)$ is empty for any field $F$ of characteristic $p$.
From basic valuation theory $^{*}$ we know that $(E_p)$ implies $(E_0)$ for any $p \geq 0$ (this is analogous to the fact that a diophantine equation which has solutions in $\mathbb{Z}$ has (obviously) solutions in each $\mathbb{F}_p$).
I'm interested in the converse implication.
It is false in general that $(E_0)$ implies $(E_p)$ (for a fixed $p>0$) : just consider the set $S$ which contains only the constant polynomial $p$. But in this counterexample $E_l$ fails for only one prime $l$.

Hence the question : is it true that if $(E_0)$ holds, then $(E_p)$ also holds for all but finitely many primes ? If yes, is there some effective upper bound (in term of $S$) for the greatest prime for which $(E_p)$ fails ?

$^{*}$ Here is the argument : Let $F$ be a field of characteristic $0$, and suppose $X(F)$ contains a line $x$ of $F^n$. Since $\mathbb{Q} \subset F$, some non-archimedean norm $ |\cdot|$ on $F$ extends the $p$-adic norm $ |\cdot|_p$ (and has the same range). If $R = \overline{B}(0,1)$ and $ \mathfrak{m} = B(0,1)$ then $K = R/\mathfrak{m}$ is a field of characteristic $p$. Then the image of $x \cap R^n$ in $K^n$ is itself a line, therefore an element of $X(K)$ : contradiction.

I probably should have also mentioned that in the simplest case $F_1(X,Y)=F_2(X,Y)=0$, the existence of a nontrivial solution in an algebraically closed field $k$ is exactly the condition that the resultant $R(F_1,F_2)$ vanishes in $k$.
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Joe SilvermanApr 22 '12 at 15:38

2 Answers
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No. For example, $3x^3+4y^3+5z^3=0$ has nontrivial solutions mod $p$ for every prime $p$, but it has no nontrivial solutions in $\mathbb{Q}$. Indeed, it has solutions in every $p$-adic field $\mathbb{Q}_p$ and also solutions in $\mathbb{R}$. This is a famous example of Selmer. One says that the "Hasse principle" fails for this example. There is a subtler obstruction to "Local Solutions for all $p$ Implies Global Solutions" called the Brauer-Manin obstruction. The vanishing of the Brauer-Manin obstruction is necessary for the existence of a solution in $\mathbb{Q}$, but it is not always a sufficient condition. The study of such obstructions is an important area of current research.

EDIT: Sorry, I misinterpreted your question. I guess you're asking whether "no solutions in $\mathbb{C}$ implies there is some prime $p$ such that no solutions in $\overline{\mathbb{F}}_p$." The answer to that should be yes. Let $F_1,\ldots,F_r$ be polynomials in your set $S$ that generate the homogeneous ideal generated by the elements of $S$. Then elimination theory says that there is a list of polynomials $E_1,\ldots,E_s$ in the coefficients of $F_1,\ldots,F_r$ such that $F_1,\ldots,F_r$ have a common solution in $\mathbb{P}^{n-1}(k)$ if and only if every $E_i$ vanishes at $F_1,\ldots,F_r$. (This is for algebraically closed fields $k$.) So if $F_1,\ldots,F_r$ have no common zeros in $\mathbb{P}^{n-1}(\mathbb{C})$, then some $E_i(F_1,\ldots,F_r)$ is non-zero, and hence it is nonzero in characteristic $p$ for all but finitely many $p$, and hence $F_1,\ldots,F_r$ have no common zeros in $\mathbb{P}^{n-1}(\overline{\mathbb{F}}_p)$ for all but finitely many $p$. For effective estimates, the term you want to search for is "effective nullstellensatz".

Here is an argument which avoids elimination theory at the cost of being highly nonconstructive. We prove the contrapositive: if $X(F)$ is non-empty for fields $F$ of infinitely many different positive characteristics, then $X(F)$ is non-empty for a field $F$ of characteristic zero.

Let $F_i$ be an infinite collection of fields of distinct characteristics such that $X(F_i)$ is non-empty and consider the ultraproduct $F = \prod F_i/U$. By Łoś's theorem $F$ has characteristic $0$, and by choosing a point in each $X(F_i)$ we obtain a point in $X(F)$ by taking coordinatewise ultraproducts.

Instead of ultraproducts, you could use the compactness theorem of first-order logic directly. Let $T$ be the theory, in the language of fields plus $n$ new constant symbols $c_i$, consisting of the axioms for fields, the equations saying that the $c_i$ are roots of all the polynomials in $S$, and the formulas $1+\dots+1\neq0$ for each prime number of summands. If $X(F)$ is nonempty for fields of infinitely many prime characteristics, then each finite subset of $T$ is satisfiable. By compactness, there is a model of all of $T$, and that's a field $F$ of characteristic 0 with nonempty $X(F)$.
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Andreas BlassApr 22 '12 at 20:32

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In the comment I just posted, $T$ should also include the formulas $c_i\neq 0$ for the $n$ new constants, so that you're talking about solutions to $S$ in projective space.
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Andreas BlassApr 22 '12 at 20:34

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Correction to the correction: $T$ should not say that all the $c_i$ are non-zero but that at least one of them is. In other words, it should contain the disjunction $c_1\neq0\lor\dots\lor c_n\neq0$.
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Andreas BlassApr 28 '12 at 16:40